Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ -x^4+3x^2+112x = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 0 & x_2 = 5.0276 \end{matrix} $$(you can use the step-by-step polynomial equation solver to see a detailed explanation of how to solve the equation)
STEP 2:Find the y-intercepts
To find the y-intercepts, substitute $ x = 0 $ into $ \color{blue}{ p(x) = -x^4+3x^2+112x } $, so:
$$ \text{Y inercept} = p(0) = 0 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( -x^4+3x^2+112x \right) = \lim_{x \to -\infty} -x^4 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( -x^4+3x^2+112x \right) = \lim_{x \to \infty} -x^4 = \color{blue}{ -\infty } $$The graph ends in the lower-right corner.
STEP 4:Find the turning points
To determine the turning point, we need to find the first derivative of $ p(x) $:
$$ p^{\prime} (x) = -4x^3+6x+112 $$The x coordinate of the turning point is located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x = 3.2011 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinate, substitute the above value into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 3.2011 } \Rightarrow p\left(3.2011\right) = \color{orangered}{ 284.2625 }\end{aligned} $$So the turning point is:
$$ \begin{matrix} \left( 3.2011, 284.2625 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -12x^2+6 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = \dfrac{\sqrt{ 2 }}{ 2 } & x_2 = - \dfrac{\sqrt{ 2 }}{ 2 } \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ \frac{\sqrt{ 2 }}{ 2 } } \Rightarrow p\left(\frac{\sqrt{ 2 }}{ 2 }\right) = \color{orangered}{ 80.446 }\\[1 em] \text{for } ~ x & = \color{blue}{ - \frac{\sqrt{ 2 }}{ 2 } } \Rightarrow p\left(- \frac{\sqrt{ 2 }}{ 2 }\right) = \color{orangered}{ -77.946 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( \dfrac{\sqrt{ 2 }}{ 2 }, 80.446 \right) & \left( - \dfrac{\sqrt{ 2 }}{ 2 }, -77.946 \right)\end{matrix} $$